I have a very basic question regarding algebraic model theory. I am trying to read Espaces de Berkovich, polytopes, squelettes et théorie des modèles (MSN) by Antoine Ducros. The relevant section is Section 0.31. Let me give the setup: Let $K$ be a valued field and let $\mathscr{C}_K$ be the category of non-trivially valued, algebraically closed field extensions of $K$. We write $\mathscr{L}_\mathrm{val}$ for the language of valued fields, having three sorts (the valued field, its residue field and the value group).

Let $\mathscr{S}$ be a product of sorts. For each model of the theory of non-trivially valued, algebraically closed field extensions of $K$, i.e. for each $F \in \mathscr{C}_K$, we can then make sense of $\mathscr{S}(F)$, and this gives a functor $\mathscr{S} \colon \mathscr{C}_K \to \mathrm{Set}$. If $\Phi(x)$ is a formula of $\mathscr{L}_\mathrm{val}$ (with parameters in $K$), whose free variables $x$ live in $\mathscr{S}$, then the subfunctor of $\mathscr{S}$ mapping $F$ to $\{x \in \mathscr{S}(F) \mid \Phi(x)\}$ is called a definable subfunctor of $\mathscr{S}$.

In general, an abstract functor from $\mathscr{C}_K$ to $\mathrm{Set}$ is called definable if it is isomorphic to a definable subfunctor of some product of sorts.

Now it is claimed that a scheme of finite type $\mathscr{X}$ over $K$ gives rise to a definable functor on $\mathscr{C}_K$. I don't understand why this is true. If $\mathscr{X}$ is affine, then I believe that the associated functor should simply be the “functor of points”. Namely, if $\mathscr{X}$ is isomorphic to $\mathrm{Spec}(K[T_1, \dots, T_n] / \langle f_1, \dots, f_r\rangle)$, then its associated functor is isomorphic to the definable subfunctor of $F \mapsto F^n$, given by the equations $f_i$. I have no idea, what is meant in the non-affine case. I would be very glad about some clarification. Also, if some of my writing above hints at some misunderstanding on my side, please let me know.


This follows by elimination of imaginaries in algebraically closed fields. Given a finite affine cover of your scheme, one obtains a functor to Set by gluing the affine charts. This is indeed not definable, but a quotient of a definable set by a definable equivalence relation. Elimination of imaginaries tells you precisely that such a quotient is in definable bijection with a definable set. You can find the details in Chapter 4 (A. Pillay, Model theory of algebraically closed fields) of the following book:

Bouscaren, Elisabeth, Introduction to model theory, Bouscaren, Elisabeth (ed.), Model theory and algebraic geometry. An introduction to E. Hrushovski’s proof of the geometric Mordell-Lang conjecture. Berlin: Springer. Lect. Notes Math. 1696, 1-18 (1998). ZBL0925.03160.

Chapters 1 and 2 also contain an introduction to general model theoretic concepts including elimination of imaginaries (Chapter 2).

  • 1
    $\begingroup$ I edited in links to the book and the individual chapters. Since you later cite Chapter 1 separately, I think you didn't mean your reference and corresponding ZBL link to refer only to the first chapter, but rather to the whole book. That's ZBL0920.03046. $\endgroup$ – LSpice Jun 6 '20 at 18:05
  • $\begingroup$ Thank you very much. $\endgroup$ – Cubikova Jun 6 '20 at 18:18
  • $\begingroup$ If I understand “gluing affine charts” correctly, then you are saying that also for general $\mathscr{X}$, the associated functor is the functor of points, sending $F$ to $\mathscr{X}(F) = \mathrm{Hom}_K(\mathrm{Spec}(F), \mathscr{X})$, right? Could you maybe indicate for some simple non-affine scheme like $\mathbf{P}^1_K$, how the bijection that you get to a definable functor would look like? $\endgroup$ – Jakob Werner Jun 6 '20 at 18:22
  • $\begingroup$ By the way, thank you for the hints to the literature, I will definitely have a look at it. $\endgroup$ – Jakob Werner Jun 6 '20 at 18:24
  • $\begingroup$ Indeed, the functor is isomorphic to the functor of points. In the case of $\mathbf{P}_K^n$, the functor of points is isomorphic to the functor sending $F$ to the set of $n$-dimensional subspaces of $F^{n+1}$. The usual equivalence relation on $F^{n+1}$ given by $(x_0,\ldots, x_n)\sim(y_0,\ldots,y_n)$ if and only if there is $\lambda\in F^*$ such that $\lambda x_i=y_i$ for each $i\in \{0,\ldots,n\}$ is the corresponding definable equivalence relation. $\endgroup$ – Cubikova Jun 8 '20 at 9:47

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